Integrand size = 21, antiderivative size = 199 \[ \int e^{\frac {x}{2}+x z} x^4 \sin ^4(\pi z) \, dz=\frac {24 e^{\frac {x}{2}+x z} \pi ^4 x^3}{64 \pi ^4+20 \pi ^2 x^2+x^4}-\frac {24 e^{\frac {x}{2}+x z} \pi ^3 x^4 \cos (\pi z) \sin (\pi z)}{64 \pi ^4+20 \pi ^2 x^2+x^4}+\frac {12 e^{\frac {x}{2}+x z} \pi ^2 x^5 \sin ^2(\pi z)}{64 \pi ^4+20 \pi ^2 x^2+x^4}-\frac {4 e^{\frac {x}{2}+x z} \pi x^4 \cos (\pi z) \sin ^3(\pi z)}{16 \pi ^2+x^2}+\frac {e^{\frac {x}{2}+x z} x^5 \sin ^4(\pi z)}{16 \pi ^2+x^2} \]
24*exp(1/2*x+x*z)*Pi^4*x^3/(64*Pi^4+20*Pi^2*x^2+x^4)-24*exp(1/2*x+x*z)*Pi^ 3*x^4*cos(Pi*z)*sin(Pi*z)/(64*Pi^4+20*Pi^2*x^2+x^4)+12*exp(1/2*x+x*z)*Pi^2 *x^5*sin(Pi*z)^2/(64*Pi^4+20*Pi^2*x^2+x^4)-4*exp(1/2*x+x*z)*Pi*x^4*cos(Pi* z)*sin(Pi*z)^3/(16*Pi^2+x^2)+exp(1/2*x+x*z)*x^5*sin(Pi*z)^4/(16*Pi^2+x^2)
Time = 0.15 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.68 \[ \int e^{\frac {x}{2}+x z} x^4 \sin ^4(\pi z) \, dz=\frac {e^{x \left (\frac {1}{2}+z\right )} x^4 \left (192 \pi ^4+60 \pi ^2 x^2+3 x^4-4 x^2 \left (16 \pi ^2+x^2\right ) \cos (2 \pi z)+x^2 \left (4 \pi ^2+x^2\right ) \cos (4 \pi z)-128 \pi ^3 x \sin (2 \pi z)-8 \pi x^3 \sin (2 \pi z)+16 \pi ^3 x \sin (4 \pi z)+4 \pi x^3 \sin (4 \pi z)\right )}{8 \left (64 \pi ^4 x+20 \pi ^2 x^3+x^5\right )} \]
(E^(x*(1/2 + z))*x^4*(192*Pi^4 + 60*Pi^2*x^2 + 3*x^4 - 4*x^2*(16*Pi^2 + x^ 2)*Cos[2*Pi*z] + x^2*(4*Pi^2 + x^2)*Cos[4*Pi*z] - 128*Pi^3*x*Sin[2*Pi*z] - 8*Pi*x^3*Sin[2*Pi*z] + 16*Pi^3*x*Sin[4*Pi*z] + 4*Pi*x^3*Sin[4*Pi*z]))/(8* (64*Pi^4*x + 20*Pi^2*x^3 + x^5))
Time = 0.38 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.90, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {27, 4934, 4934, 2624}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int x^4 e^{x z+\frac {x}{2}} \sin ^4(\pi z) \, dz\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle x^4 \int e^{z x+\frac {x}{2}} \sin ^4(\pi z)dz\) |
| \(\Big \downarrow \) 4934 |
| \(\displaystyle x^4 \left (\frac {12 \pi ^2 \int e^{z x+\frac {x}{2}} \sin ^2(\pi z)dz}{x^2+16 \pi ^2}+\frac {x e^{x z+\frac {x}{2}} \sin ^4(\pi z)}{x^2+16 \pi ^2}-\frac {4 \pi e^{x z+\frac {x}{2}} \sin ^3(\pi z) \cos (\pi z)}{x^2+16 \pi ^2}\right )\) |
| \(\Big \downarrow \) 4934 |
| \(\displaystyle x^4 \left (\frac {12 \pi ^2 \left (\frac {2 \pi ^2 \int e^{z x+\frac {x}{2}}dz}{x^2+4 \pi ^2}+\frac {x e^{x z+\frac {x}{2}} \sin ^2(\pi z)}{x^2+4 \pi ^2}-\frac {2 \pi e^{x z+\frac {x}{2}} \sin (\pi z) \cos (\pi z)}{x^2+4 \pi ^2}\right )}{x^2+16 \pi ^2}+\frac {x e^{x z+\frac {x}{2}} \sin ^4(\pi z)}{x^2+16 \pi ^2}-\frac {4 \pi e^{x z+\frac {x}{2}} \sin ^3(\pi z) \cos (\pi z)}{x^2+16 \pi ^2}\right )\) |
| \(\Big \downarrow \) 2624 |
| \(\displaystyle x^4 \left (\frac {x e^{x z+\frac {x}{2}} \sin ^4(\pi z)}{x^2+16 \pi ^2}-\frac {4 \pi e^{x z+\frac {x}{2}} \sin ^3(\pi z) \cos (\pi z)}{x^2+16 \pi ^2}+\frac {12 \pi ^2 \left (\frac {2 \pi ^2 e^{x z+\frac {x}{2}}}{x \left (x^2+4 \pi ^2\right )}+\frac {x e^{x z+\frac {x}{2}} \sin ^2(\pi z)}{x^2+4 \pi ^2}-\frac {2 \pi e^{x z+\frac {x}{2}} \sin (\pi z) \cos (\pi z)}{x^2+4 \pi ^2}\right )}{x^2+16 \pi ^2}\right )\) |
x^4*((-4*E^(x/2 + x*z)*Pi*Cos[Pi*z]*Sin[Pi*z]^3)/(16*Pi^2 + x^2) + (E^(x/2 + x*z)*x*Sin[Pi*z]^4)/(16*Pi^2 + x^2) + (12*Pi^2*((2*E^(x/2 + x*z)*Pi^2)/ (x*(4*Pi^2 + x^2)) - (2*E^(x/2 + x*z)*Pi*Cos[Pi*z]*Sin[Pi*z])/(4*Pi^2 + x^ 2) + (E^(x/2 + x*z)*x*Sin[Pi*z]^2)/(4*Pi^2 + x^2)))/(16*Pi^2 + x^2))
3.3.75.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((F_)^(v_))^(n_.), x_Symbol] :> Simp[(F^v)^n/(n*Log[F]*D[v, x]), x] /; FreeQ[{F, n}, x] && LinearQ[v, x]
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)]^(n_), x_Symbo l] :> Simp[b*c*Log[F]*F^(c*(a + b*x))*(Sin[d + e*x]^n/(e^2*n^2 + b^2*c^2*Lo g[F]^2)), x] + (-Simp[e*n*F^(c*(a + b*x))*Cos[d + e*x]*(Sin[d + e*x]^(n - 1 )/(e^2*n^2 + b^2*c^2*Log[F]^2)), x] + Simp[(n*(n - 1)*e^2)/(e^2*n^2 + b^2*c ^2*Log[F]^2) Int[F^(c*(a + b*x))*Sin[d + e*x]^(n - 2), x], x]) /; FreeQ[{ F, a, b, c, d, e}, x] && NeQ[e^2*n^2 + b^2*c^2*Log[F]^2, 0] && GtQ[n, 1]
Time = 1.58 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.63
| method | result | size |
| default | \(x^{4} \left (\frac {3 \,{\mathrm e}^{x \left (\frac {1}{2}+z \right )}}{8 x}+\frac {x \,{\mathrm e}^{\frac {1}{2} x +x z} \cos \left (4 \pi z \right )}{128 \pi ^{2}+8 x^{2}}+\frac {\pi \,{\mathrm e}^{\frac {1}{2} x +x z} \sin \left (4 \pi z \right )}{32 \pi ^{2}+2 x^{2}}-\frac {x \,{\mathrm e}^{\frac {1}{2} x +x z} \cos \left (2 \pi z \right )}{2 \left (4 \pi ^{2}+x^{2}\right )}-\frac {\pi \,{\mathrm e}^{\frac {1}{2} x +x z} \sin \left (2 \pi z \right )}{4 \pi ^{2}+x^{2}}\right )\) | \(125\) |
| risch | \(\frac {3 x^{3} {\mathrm e}^{\frac {x \left (1+2 z \right )}{2}}}{8}+\frac {x^{5} {\mathrm e}^{\frac {1}{2} x +x z} \cos \left (4 \pi z \right )}{128 \pi ^{2}+8 x^{2}}+\frac {x^{4} {\mathrm e}^{\frac {1}{2} x +x z} \pi \sin \left (4 \pi z \right )}{32 \pi ^{2}+2 x^{2}}-\frac {x^{5} {\mathrm e}^{\frac {1}{2} x +x z} \cos \left (2 \pi z \right )}{2 \left (4 \pi ^{2}+x^{2}\right )}-\frac {x^{4} {\mathrm e}^{\frac {1}{2} x +x z} \pi \sin \left (2 \pi z \right )}{4 \pi ^{2}+x^{2}}\) | \(134\) |
| parallelrisch | \(\frac {x^{3} {\mathrm e}^{\frac {x \left (1+2 z \right )}{2}} \left (-128 \pi ^{3} x \sin \left (2 \pi z \right )+16 \pi ^{3} x \sin \left (4 \pi z \right )+4 \pi ^{2} x^{2} \cos \left (4 \pi z \right )-64 \pi ^{2} x^{2} \cos \left (2 \pi z \right )-8 x^{3} \pi \sin \left (2 \pi z \right )+4 x^{3} \pi \sin \left (4 \pi z \right )+x^{4} \cos \left (4 \pi z \right )-4 x^{4} \cos \left (2 \pi z \right )+192 \pi ^{4}+60 \pi ^{2} x^{2}+3 x^{4}\right )}{512 \pi ^{4}+160 \pi ^{2} x^{2}+8 x^{4}}\) | \(142\) |
| norman | \(\frac {\frac {24 \,{\mathrm e}^{\frac {1}{2} x +x z} \pi ^{4} x^{3}}{64 \pi ^{4}+20 \pi ^{2} x^{2}+x^{4}}-\frac {48 \pi ^{3} x^{4} {\mathrm e}^{\frac {1}{2} x +x z} \tan \left (\frac {\pi z}{2}\right )}{64 \pi ^{4}+20 \pi ^{2} x^{2}+x^{4}}+\frac {48 \pi ^{3} x^{4} {\mathrm e}^{\frac {1}{2} x +x z} \left (\tan ^{7}\left (\frac {\pi z}{2}\right )\right )}{64 \pi ^{4}+20 \pi ^{2} x^{2}+x^{4}}+\frac {16 \left (9 \pi ^{4}+10 \pi ^{2} x^{2}+x^{4}\right ) x^{3} {\mathrm e}^{\frac {1}{2} x +x z} \left (\tan ^{4}\left (\frac {\pi z}{2}\right )\right )}{64 \pi ^{4}+20 \pi ^{2} x^{2}+x^{4}}+\frac {24 \,{\mathrm e}^{\frac {1}{2} x +x z} \pi ^{4} x^{3} \left (\tan ^{8}\left (\frac {\pi z}{2}\right )\right )}{64 \pi ^{4}+20 \pi ^{2} x^{2}+x^{4}}-\frac {16 \pi \,x^{4} \left (11 \pi ^{2}+2 x^{2}\right ) {\mathrm e}^{\frac {1}{2} x +x z} \left (\tan ^{3}\left (\frac {\pi z}{2}\right )\right )}{64 \pi ^{4}+20 \pi ^{2} x^{2}+x^{4}}+\frac {16 \pi \,x^{4} \left (11 \pi ^{2}+2 x^{2}\right ) {\mathrm e}^{\frac {1}{2} x +x z} \left (\tan ^{5}\left (\frac {\pi z}{2}\right )\right )}{64 \pi ^{4}+20 \pi ^{2} x^{2}+x^{4}}+\frac {48 x^{3} \left (2 \pi ^{2}+x^{2}\right ) \pi ^{2} {\mathrm e}^{\frac {1}{2} x +x z} \left (\tan ^{2}\left (\frac {\pi z}{2}\right )\right )}{64 \pi ^{4}+20 \pi ^{2} x^{2}+x^{4}}+\frac {48 x^{3} \left (2 \pi ^{2}+x^{2}\right ) \pi ^{2} {\mathrm e}^{\frac {1}{2} x +x z} \left (\tan ^{6}\left (\frac {\pi z}{2}\right )\right )}{64 \pi ^{4}+20 \pi ^{2} x^{2}+x^{4}}}{\left (1+\tan ^{2}\left (\frac {\pi z}{2}\right )\right )^{4}}\) | \(433\) |
x^4*(3/8*exp(x*(1/2+z))/x+1/8*x/(16*Pi^2+x^2)*exp(1/2*x+x*z)*cos(4*Pi*z)+1 /2*Pi/(16*Pi^2+x^2)*exp(1/2*x+x*z)*sin(4*Pi*z)-1/2*x/(4*Pi^2+x^2)*exp(1/2* x+x*z)*cos(2*Pi*z)-Pi/(4*Pi^2+x^2)*exp(1/2*x+x*z)*sin(2*Pi*z))
Time = 0.28 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.73 \[ \int e^{\frac {x}{2}+x z} x^4 \sin ^4(\pi z) \, dz=\frac {4 \, {\left ({\left (4 \, \pi ^{3} x^{4} + \pi x^{6}\right )} \cos \left (\pi z\right )^{3} - {\left (10 \, \pi ^{3} x^{4} + \pi x^{6}\right )} \cos \left (\pi z\right )\right )} e^{\left (x z + \frac {1}{2} \, x\right )} \sin \left (\pi z\right ) + {\left (24 \, \pi ^{4} x^{3} + 16 \, \pi ^{2} x^{5} + x^{7} + {\left (4 \, \pi ^{2} x^{5} + x^{7}\right )} \cos \left (\pi z\right )^{4} - 2 \, {\left (10 \, \pi ^{2} x^{5} + x^{7}\right )} \cos \left (\pi z\right )^{2}\right )} e^{\left (x z + \frac {1}{2} \, x\right )}}{64 \, \pi ^{4} + 20 \, \pi ^{2} x^{2} + x^{4}} \]
(4*((4*pi^3*x^4 + pi*x^6)*cos(pi*z)^3 - (10*pi^3*x^4 + pi*x^6)*cos(pi*z))* e^(x*z + 1/2*x)*sin(pi*z) + (24*pi^4*x^3 + 16*pi^2*x^5 + x^7 + (4*pi^2*x^5 + x^7)*cos(pi*z)^4 - 2*(10*pi^2*x^5 + x^7)*cos(pi*z)^2)*e^(x*z + 1/2*x))/ (64*pi^4 + 20*pi^2*x^2 + x^4)
Time = 151.38 (sec) , antiderivative size = 1277, normalized size of antiderivative = 6.42 \[ \int e^{\frac {x}{2}+x z} x^4 \sin ^4(\pi z) \, dz=\text {Too large to display} \]
x**4*Piecewise((3*z*sin(pi*z)**4/8 + 3*z*sin(pi*z)**2*cos(pi*z)**2/4 + 3*z *cos(pi*z)**4/8 - 5*sin(pi*z)**3*cos(pi*z)/(8*pi) - 3*sin(pi*z)*cos(pi*z)* *3/(8*pi), Eq(x, 0)), (z*exp(-4*I*pi*z)*sin(pi*z)**4/16 - I*z*exp(-4*I*pi* z)*sin(pi*z)**3*cos(pi*z)/4 - 3*z*exp(-4*I*pi*z)*sin(pi*z)**2*cos(pi*z)**2 /8 + I*z*exp(-4*I*pi*z)*sin(pi*z)*cos(pi*z)**3/4 + z*exp(-4*I*pi*z)*cos(pi *z)**4/16 + 7*I*exp(-4*I*pi*z)*sin(pi*z)**4/(24*pi) + 11*exp(-4*I*pi*z)*si n(pi*z)**3*cos(pi*z)/(48*pi) + 5*exp(-4*I*pi*z)*sin(pi*z)*cos(pi*z)**3/(48 *pi) - I*exp(-4*I*pi*z)*cos(pi*z)**4/(24*pi), Eq(x, -4*I*pi)), (-z*exp(-2* I*pi*z)*sin(pi*z)**4/4 + I*z*exp(-2*I*pi*z)*sin(pi*z)**3*cos(pi*z)/2 + I*z *exp(-2*I*pi*z)*sin(pi*z)*cos(pi*z)**3/2 + z*exp(-2*I*pi*z)*cos(pi*z)**4/4 - 5*I*exp(-2*I*pi*z)*sin(pi*z)**4/(24*pi) + exp(-2*I*pi*z)*sin(pi*z)**3*c os(pi*z)/(3*pi) - I*exp(-2*I*pi*z)*sin(pi*z)**2*cos(pi*z)**2/(2*pi) - I*ex p(-2*I*pi*z)*cos(pi*z)**4/(8*pi), Eq(x, -2*I*pi)), (-z*exp(2*I*pi*z)*sin(p i*z)**4/4 - I*z*exp(2*I*pi*z)*sin(pi*z)**3*cos(pi*z)/2 - I*z*exp(2*I*pi*z) *sin(pi*z)*cos(pi*z)**3/2 + z*exp(2*I*pi*z)*cos(pi*z)**4/4 + 5*I*exp(2*I*p i*z)*sin(pi*z)**4/(24*pi) + exp(2*I*pi*z)*sin(pi*z)**3*cos(pi*z)/(3*pi) + I*exp(2*I*pi*z)*sin(pi*z)**2*cos(pi*z)**2/(2*pi) + I*exp(2*I*pi*z)*cos(pi* z)**4/(8*pi), Eq(x, 2*I*pi)), (z*exp(4*I*pi*z)*sin(pi*z)**4/16 + I*z*exp(4 *I*pi*z)*sin(pi*z)**3*cos(pi*z)/4 - 3*z*exp(4*I*pi*z)*sin(pi*z)**2*cos(pi* z)**2/8 - I*z*exp(4*I*pi*z)*sin(pi*z)*cos(pi*z)**3/4 + z*exp(4*I*pi*z)*...
Time = 0.23 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.80 \[ \int e^{\frac {x}{2}+x z} x^4 \sin ^4(\pi z) \, dz=\frac {{\left ({\left (4 \, \pi ^{2} x^{2} + x^{4}\right )} \cos \left (4 \, \pi z\right ) e^{\left (x z + \frac {1}{2} \, x\right )} - 4 \, {\left (16 \, \pi ^{2} x^{2} + x^{4}\right )} \cos \left (2 \, \pi z\right ) e^{\left (x z + \frac {1}{2} \, x\right )} + 4 \, {\left (4 \, \pi ^{3} x + \pi x^{3}\right )} e^{\left (x z + \frac {1}{2} \, x\right )} \sin \left (4 \, \pi z\right ) - 8 \, {\left (16 \, \pi ^{3} x + \pi x^{3}\right )} e^{\left (x z + \frac {1}{2} \, x\right )} \sin \left (2 \, \pi z\right ) + 3 \, {\left (64 \, \pi ^{4} + 20 \, \pi ^{2} x^{2} + x^{4}\right )} e^{\left (x z + \frac {1}{2} \, x\right )}\right )} x^{4}}{8 \, {\left (64 \, \pi ^{4} x + 20 \, \pi ^{2} x^{3} + x^{5}\right )}} \]
1/8*((4*pi^2*x^2 + x^4)*cos(4*pi*z)*e^(x*z + 1/2*x) - 4*(16*pi^2*x^2 + x^4 )*cos(2*pi*z)*e^(x*z + 1/2*x) + 4*(4*pi^3*x + pi*x^3)*e^(x*z + 1/2*x)*sin( 4*pi*z) - 8*(16*pi^3*x + pi*x^3)*e^(x*z + 1/2*x)*sin(2*pi*z) + 3*(64*pi^4 + 20*pi^2*x^2 + x^4)*e^(x*z + 1/2*x))*x^4/(64*pi^4*x + 20*pi^2*x^3 + x^5)
Time = 0.27 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.57 \[ \int e^{\frac {x}{2}+x z} x^4 \sin ^4(\pi z) \, dz=\frac {1}{8} \, {\left ({\left (\frac {x \cos \left (4 \, \pi z\right )}{16 \, \pi ^{2} + x^{2}} + \frac {4 \, \pi \sin \left (4 \, \pi z\right )}{16 \, \pi ^{2} + x^{2}}\right )} e^{\left (x z + \frac {1}{2} \, x\right )} - 4 \, {\left (\frac {x \cos \left (2 \, \pi z\right )}{4 \, \pi ^{2} + x^{2}} + \frac {2 \, \pi \sin \left (2 \, \pi z\right )}{4 \, \pi ^{2} + x^{2}}\right )} e^{\left (x z + \frac {1}{2} \, x\right )} + \frac {3 \, e^{\left (x z + \frac {1}{2} \, x\right )}}{x}\right )} x^{4} \]
1/8*((x*cos(4*pi*z)/(16*pi^2 + x^2) + 4*pi*sin(4*pi*z)/(16*pi^2 + x^2))*e^ (x*z + 1/2*x) - 4*(x*cos(2*pi*z)/(4*pi^2 + x^2) + 2*pi*sin(2*pi*z)/(4*pi^2 + x^2))*e^(x*z + 1/2*x) + 3*e^(x*z + 1/2*x)/x)*x^4
Time = 1.12 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.70 \[ \int e^{\frac {x}{2}+x z} x^4 \sin ^4(\pi z) \, dz=\frac {x^3\,{\mathrm {e}}^{\frac {x}{2}+x\,z}\,\left (24\,\Pi ^4-\frac {x^4\,\cos \left (2\,\Pi \,z\right )}{2}+\frac {x^4\,\cos \left (4\,\Pi \,z\right )}{8}+\frac {3\,x^4}{8}+\frac {15\,\Pi ^2\,x^2}{2}-\Pi \,x^3\,\sin \left (2\,\Pi \,z\right )-16\,\Pi ^3\,x\,\sin \left (2\,\Pi \,z\right )+\frac {\Pi \,x^3\,\sin \left (4\,\Pi \,z\right )}{2}+2\,\Pi ^3\,x\,\sin \left (4\,\Pi \,z\right )-8\,\Pi ^2\,x^2\,\cos \left (2\,\Pi \,z\right )+\frac {\Pi ^2\,x^2\,\cos \left (4\,\Pi \,z\right )}{2}\right )}{64\,\Pi ^4+20\,\Pi ^2\,x^2+x^4} \]